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2 years ago

Competitive inhibition occurs when a

Chemistry

2 answers:

Aleksandr-060686 [28]2 years ago

6 0

The answer to your question is C.

ioda2 years ago

3 0

Answer:

Competitive inhibition occurs when a molecule binds to an enzyme in the active site and prevents the substrate from binding. (C)

Explanation:

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WHAT IT IZ BRAINLY anyways I got a question for y’all do you think someone can forget who they rlly loved ? Can you ever forget

alexdok [17]

Answer:

Its a 50% chance of it happeing

Explanation:

5 0

3 years ago

When the glucose solution described in part A is connected to an unknown solution via a semipermeable membrane, the unknown solu

olya-2409 [2.1K]

Answer: The unknown solution had the lower concentration

Explanation: concentration will always move from higher to lower region. If the concentration of the unknown solution has increased, it therefore means that the initial concentration of the unknown solution was low

7 0

4 years ago

A 2.04 g lead weight, initially at 10.8 oC, is submerged in 7.62 g of water at 52.3 oC in an insulated container. clear = 0.128

alisha [4.7K]

Answer: The final temperature of both the weight and the water at thermal equilibrium is $50.26^{o}C$ .

Explanation:

The given data is as follows.

mass = 7.62 g, $T_{2} = 10.8^{o}C$

Let us assume that T be the final temperature. Therefore, heat lost by water is calculated as follows.

q = $mC \times \Delta T$

= $7.62 g \times 4.184 J/^{o}C \times (52.3 - T)$

Now, heat gained by lead will be calculated as follows.

q = $mC \times \text{Temperature change of lead}$

= $2.04 \times 0.128 \times (T - 11.0)$

According to the given situation,

Heat lost = Heat gained

$7.62 g \times 4.184 J/^{o}C \times (52.3 - T)$ = $2.04 \times 0.128 \times (T - 11.0)$

T = $50.26^{o}C$

Thus, we can conclude that the final temperature of both the weight and the water at thermal equilibrium is $50.26^{o}C$ .

7 0

3 years ago

Choose the aqueous solution below with the highest boiling point. These are all solutions of nonvolatile solutes and you should

o-na [289]

The question is incomplete, the complete question is;

Choose the aqueous solution that has the highest boiling point. These are all solutions of nonvolatile solutes and you should assume ideal van't Hoff factors where applicable. 0.100 m C6H12O6 0.100 m AlCl3 0.100 m NaCl 0.100 m MgCl2 They all have the same boiling point.

Answer:

AlCl3 0.100 m

Explanation:

Let us remember that the boiling point elevation is given by;

ΔTb = Kb m i

Where;

ΔTb = boiling point elevation

Kb = boiling point constant

m = molality of the solution

i = Van't Hoff factor

We can see from the question that all the solutions possess the same molality, ΔTb now depends on the value of the Van't Hoff factor which in turn depends on the number of particles in solution.

AlCl3 yields four particles in solution, hence ΔTb is highest for AlCl3 . The solution having the highest value of ΔTb also has the highest boiling point.

8 0

3 years ago

[H2O] = 0.077 M

salantis [7]

Cl₂O + H₂O ⇄ 2HClO

K = [HClO]²/[Cl₂O][H₂O]
K = (0,023)²/(0,077×0,077)
K = 0,000529/0,005929
K ≈ 0,0892

:)

7 0

3 years ago